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3 years ago

Select the single best answer. Determine the following type of reaction: CH3―CH2―CH(Br)―CH3 CH3―CH═CH―CH3 + NaBr + H2O

Chemistry

1 answer:

Setler79 [48]3 years ago

3 0

Answer:

Deshydrohalogenation

Explanation:

You are not providing options to answer, however, this can be answered without options.

Now, in the reaction we can see that we have an atom of Bromine in carbon 2, and in the product appears as NaBr. This means that the Br was substracted by elimination. It's an elimination because the final product do not have a substituent where the bromine was, (Like another nucleophyle such OH or another halide). If you look closely the final product, we can see that one hydrogen in carbon 3, is no longer there. So this electrophyle was also substracted, in this case, by a base (Such NaOH), so in this case, it's ocurring an elimination reaction via E2 (One step, bimolecular). So, as the final product has been substracted the nucleophyle and electrophyle, this treaction is a deshydrohalogenation (an atom of hydrogen and a halide were substracted). The mechanism of this, you can see it in the picture.

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Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

WHAT IS THE PERCENT BY VOLUME OF ETHANOL IN A SOLUTION THAT CONTAINS 35 mL ETHANOL IN 115 mL OF WATER?

Dmitry [639]

We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.

3 0

3 years ago

If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? ball A,

rewona [7]

Density (D) is defined as the amount or mass (m) of a substance present in a unit volume(V). It can be expressed mathematically as:

Density = Mass/Volume

i.e. D = m/V -------(1)

Units: g/cm3

If m1, V1 and D1 = mass, volume and density respectively of ball A

m2, V2 and D2 = mass, volume and density respectively of ball B

It is given that: V1 = V2; but m1 = 2m2

Based on equation (1) we have:

D1/D2 = (m1/V1)* (V2/m2) = (2m2/V2)*(V2/m2) = 2

Thus, density of ball A is twice that of B.

Ans: Ball A will have a greater density than B

If m1, V1 and D1 = mass, volume and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

It is given that:

V1 = 3V2 and m2 = 1/3(m1) i.e m1 = 3m2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (3m2/3V2)*(V2/m2) = 1

Thus, D1 = D2

Ans: Their densities are equal

If m1, V1 and D1 = mass, volume and density respectively of ball P

m2, V2 and D2 = mass, volume and density respectively of ball Q

It is given that:

m1 = m2 but V1 = 2V2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (m2/2V2)*(V2/m2) = 1/2

Thus, D1 = 1/2(D2)

Ans: Ball Q will have a greater density.

If m1, V1 and D1 = mass, volume and density respectively of ball X

m2, V2 and D2 = mass, volume and density respectively of ball Y

It is given that:

V1 = 2V2 and m1 = 1/2(m2)

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = ((1/2(m2)/2V2)*(V2/m2) = 1/4

Thus, D1 = 1/4(D2)

Ans: Ball Y will have a greater density.

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3 years ago

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AveGali [126]

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